hdu 4430 Yukari's Birthday(二分+枚举)

Yukari's Birthday

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 883    Accepted Submission(s): 151

Problem Description

Today is Yukari's n-th birthday. Ran and Chen hold a celebration party for her. Now comes the most important part, birthday cake! But it's a big challenge for them to place n candles on the top of the cake. As Yukari has lived for such a long long time, though she herself insists that she is a 17-year-old girl.
To make the birthday cake look more beautiful, Ran and Chen decide to place them like r ≥ 1 concentric circles. They place k ⁱ candles equidistantly on the i-th circle, where k ≥ 2, 1 ≤ i ≤ r. And it's optional to place at most one candle at the center of the cake. In case that there are a lot of different pairs of r and k satisfying these restrictions, they want to minimize r × k. If there is still a tie, minimize r.

 

Input

There are about 10,000 test cases. Process to the end of file.
Each test consists of only an integer 18 ≤ n ≤ 10 ¹².

 

Output

For each test case, output r and k.

 

Sample Input

   
   
   
   

    
    
    
    18
111
1111
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1 17
2 10
3 10
   
   
   
   

 

Source

2012 Asia ChangChun Regional Contest

 

Recommend

zhuyuanchen520

思路：本题数据是10^12用二分就行了，最多枚举不会超过64位，因为2^64>>10^12，因此枚举某个数的次数，看看是否有满足条件的r和k即可。

失误点：中心点可以放也可以不放。

#include<iostream>
using namespace std;
const long long oo=1000001;
///二分查找是否有满足条件的k
long long bserch(int x,long long n)
{
  if(x==1)return n;
  long long l=2,r=oo,mid,sum,bit;
  while(l<=r)
  {
    mid=(l+r)/2;sum=0;bit=1;
    for(int i=0;i<x;i++)
    {bit*=mid;sum+=bit;
     if(sum>n)
     { sum=oo*oo;
       break;
     }
    }
    if(sum==n)return mid;
    else if(sum>n)r=mid-1;
    else l=mid+1;
  }
  return 0;
}
int main()
{ long long r,k,ar,ak,n;
  while(cin>>n)
  { ar=ak=oo;
    for(int i=1;i<64;i++)
    {
      k=bserch(i,n);
      if(k!=0)
      {
        r=i;
        if(r*k<ar*ak)
          ar=r,ak=k;
        else if(r*k==ar*ak&&r<ar)
          ar=r,ak=k;
      }
      k=bserch(i,n-1);
      if(k!=0)
      {
        r=i;
        if(r*k<ar*ak)
          ar=r,ak=k;
        else if(r*k==ar*ak&&r<ar)
          ar=r,ak=k;
      }
    }
    cout<<ar<<" "<<ak<<"\n";
  }
}